Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > mat, pom;
vector<string> sol, msol;
void addintos(string &s, int a) {
  string drugi;
  do {
    char c = a % 10 + 48;
    drugi.push_back(c);
    a /= 10;
  } while (a);
  reverse(drugi.begin(), drugi.end());
  for (int i = 0; i < drugi.size(); i++) s.push_back(drugi[i]);
}
int main() {
  int n, m, maxs = -1, sl = -1;
  cin >> n >> m;
  vector<int> vi(m, 0);
  mat.insert(mat.begin(), n, vi);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> mat[i][j], maxs = max(maxs, mat[i][j]);
  for (int i = 0; i <= maxs; i++) {
    bool b = 1, b2 = 1;
    pom.clear();
    pom = mat;
    sol.clear();
    for (int j = 0; j < m; j++) {
      pom[0][j] -= i;
      if (pom[0][j] < 0) {
        b = 0;
        break;
      }
    }
    if (!b) break;
    for (int j = 0; j < i; j++) sol.push_back("row 1");
    for (int j = 0; j < m; j++) {
      int f = pom[0][j];
      if (!f) continue;
      for (int i1 = 0; i1 < n; i1++) {
        if (pom[i1][j] < f) {
          b = 0;
          break;
        }
      }
      if (b) {
        for (int i1 = 0; i1 < n; i1++) pom[i1][j] -= f;
        string a = "col ";
        addintos(a, j + 1);
        for (int i1 = 0; i1 < f; i1++) sol.push_back(a);
      } else
        break;
    }
    if (!b) continue;
    for (int i1 = 1; i1 < n; i1++) {
      int f = pom[i1][0];
      for (int j = 0; j < m; j++) {
        pom[i1][j] -= f;
        if (pom[i1][j] != 0) {
          b = 0;
          break;
        }
      }
      if (!b) break;
      string a = "row ";
      addintos(a, i1 + 1);
      for (int i1 = 0; i1 < f; i1++) sol.push_back(a);
    }
    if (!b) continue;
    if (sl < 0)
      sl = sol.size(), msol = sol;
    else if (sl > sol.size())
      sl = sol.size(), msol = sol;
  }
  cout << sl << endl;
  for (int i = 0; i < sl; i++) cout << msol[i] << endl;
  return 0;
}